walker wrote:

d.

let x - a single person, 0 - a person of couples. 1 - a person of first couples, 1 - a person of second couples

x0000 16 cases to seat together at least one couple. (x1122 - 4,x2112 - 4,x2211 - 4, 1221 - 4)

0x000 16 cases to seat together at least one couple.

00x00 8 cases to seat together at least one couple.

000x0 and 0000x 16*2 due to symmetry.

Therefore,

p=1-(16*4+8)/5!=1-8*9/5*4*3*2=1-3/5=2/5

But I've wasted a lot of time to find the way.

are there any shortcuts?

Here my take.. seems short to me.

Consider first couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl)

Consider second couple as a unit and so sits together. The # of arrangements 2*4! (2 ways for couple, 4! for the couple as a unit and the remaining 3 ppl).

Consider each couple as a unit. The # of arrangements 2*2*3! (2 ways for each couple and 3! for couples a units and the loner)

So total number of ways in which atleast one couple sits together is 2*4! + 2*4! - 2*2*3! (AUB = A+B-AB)

Total number of possible ways of seating 5!

Prob that no couple sits togehter = 1 - (4*4! - 4*3!)/5! = 2/5